UNIT 2, LESSON 5 DESCARTES LAW OF SIGNS. FINDING POLYNOMIALS WITH GIVEN ZEROS If we are given the...

UNIT 2, LESSON 5

DESCARTES LAW OF SIGNS

FINDING POLYNOMIALS WITH GIVEN ZEROS

• If we are given the zeros of a polynomial, we can generate the polynomial by first creating the factors of the polynomial.


• If we are given the zeros of a polynomial, we can generate the polynomial by first creating the factors of the polynomial.• For example, if we have the zeros 1 and 3, we

can begin by using those zeros to write the factors of our polynomial:


• If we are given the zeros of a polynomial, we can generate the polynomial by first creating the factors of the polynomial.• For example, if we have the zeros 1 and 3, we

can begin by using those zeros to write the factors of our polynomial:

31 xxaxf n


• The coefficient is there to remind us that there are infinite polynomials with these zeros. For now, let’s assume the coefficient is 1.

34

33

31

2

2

xxxf

xxxxf

xxaxf n

COMPLEX ROOTS

• Find a function having zeros 1, 3i, and -3i.

COMPLEX ROOTS


ixixxxf 331

COMPLEX ROOTS


ixixxxf 331

22 91 ixxxf

COMPLEX ROOTS


ixixxxf 331

22 91 ixxxf

91 2 xxxf

COMPLEX ROOTS


ixixxxf 331

22 91 ixxxf

91 2 xxxf

9923 xxxxf

MULTIPLICITY

• Find a polynomial with -1 as a zero of multiplicity 3, 4 as a zero of multiplicity 1, and 0 as a zero of multiplicity 1.

MULTIPLICITY

• Find a polynomial with -1 as a zero of multiplicity 3, 4 as a zero of multiplicity 1, and 0 as a zero of multiplicity 1.• Set up the factors:

xxxxf 41 3

MULTIPLICITY


xxxxf 41 3 xxxxxxf 4133 23

MULTIPLICITY


xxxxf 41 3 xxxxxxf 4133 23

xxxxxxf 4119 234

xxxxxxf 4119 2345

MORE ABOUT COMPLEX ROOTS

• If a complex number is a zero of a polynomial, then its conjugate is also a zero of the polynomial.


• If a complex number is a zero of a polynomial, then its conjugate is also a zero of the polynomial.• This is also true of rational zeros.


• Find a polynomial function with zeros and .i21

21



21

21212121 xxixixxf



21

21212121 xxixixxf

iixixix 212121212



21

21212121 xxixixxf

iixixix 212121212

22 4122 iixxixxx



21

21212121 xxixixxf

iixixix 212121212

22 4122 iixxixxx 4122 xx



21

21212121 xxixixxf

iixixix 212121212


522 xx



21

21212121 xxixixxf

iixixix 212121212


522 xx

212121212 xxx



21

21212121 xxixixxf

iixixix 212121212


522 xx

212121212 xxx

21222 xxxxx



21

21212121 xxixixxf

iixixix 212121212


522 xx

212121212 xxx

21222 xxxxx

122 xx



21

21212121 xxixixxf

1252 22 xxxxxf



21

21212121 xxixixxf

1252 22 xxxxxf

52104252 223234 xxxxxxxxxf



21

21212121 xxixixxf

1252 22 xxxxxf

52104252 223234 xxxxxxxxxf

5884 234 xxxxxf

RATIONAL ZEROS THEOREM

• Let where all coefficients are integers. If p/q is a zero of P(x), then p is a factor of a0 and q is a factor of an.

011

1 ... axaxaxaxP nn

nn


• Factor into linear factors.

410113 34 xxxxf



410113 34 xxxxf

3,1

4,2,1

q

p



410113 34 xxxxf

3,1

4,2,1

q

p

34

34

32

32

31

31 ,,,,,,4,4,2,2,1,1

q

pPossible

DESCARTES' RULE OF SIGNS

• The number of positive roots of a polynomial with real coefficients is equal to the number of "changes of sign" in the list of coefficients, or is less than this number by a multiple of 2.

http://www.cut-the-knot.org/fta/ROS2.shtml

HOW MANY OF THE ROOTS ARE POSITIVE?

http://www.mathsisfun.com/algebra/polynomials-rule-signs.html

First, rewrite the polynomial from highest to lowest exponent (ignore any "zero" terms, so it does not matter that x4 and x3 are missing):

-3x5 + x2 + 4x – 2

Then, count how many times there is a change of sign (from plus to minus, or minus to plus):

HOW MANY OF THE ROOTS ARE POSITIVE?


The number of sign changes is the maximum number of positive roots

There are 2 changes in sign, so there are at most 2 positive roots (maybe less).

HOW MANY OF THE ROOTS ARE NEGATIVE?


By doing a similar calculation we can find out how many roots are negative ...... but first we need to put "-x" in place of "x", like this:

HOW MANY OF THE ROOTS ARE NEGATIVE?


One change only, so there is 1 negative root.

HOW MANY ROOTS IN TOTAL?


The Fundamental Theorem of Algebra states that a polynomial will have exactly as many roots as its degree.

OK, we have gathered lots of info. We know all this:positive roots: 2, or 0negative roots: 1total number of roots: 5

So, after a little thought, the overall result is:•5 roots: 2 positive, 1 negative, 2 complex (one pair), or•5 roots: 0 positive, 1 negative, 4 complex (two pairs)

Consider the polynomial 3x6 - 5x4 + 2x3 - 7x2 - x + 2.

http://www.mathopolis.com/questions/q.php?id=1131


6 total roots



6 total roots4 sign changes, so 4, 2, or 0 positive roots




Now replace x by -x:⇒ 3(-x)6 - 5(-x)4 + 2(-x)3 - 7(-x)2 - (-x) + 2= 3x6 - 5x4 - 2x3 - 7x2 + x + 2




Now replace x by -x:⇒ 3(-x)6 - 5(-x)4 + 2(-x)3 - 7(-x)2 - (-x) + 2= 3x6 - 5x4 - 2x3 - 7x2 + x + 2

2 sign changes, so 2 or 0 negative roots


Consider the polynomial 2x + 5x4 - 3x2 - 7x6



First write the terms in the correct order:2x + 5x4 - 3x2 - 7x6 = -7x6 + 5x4 - 3x2 + 2x




Next factor out x :x(-7x5 + 5x3 - 3x + 2)This means x = 0 is one of the roots.




Next factor out x :x(-7x5 + 5x3 - 3x + 2)This means x = 0 is one of the roots.

There are 3 changes of sign. So the number of positive roots is either 3 or 1.



Now replace x by -x:⇒ -7(-x)5 + 5(-x)3 - 3(-x) + 2= 7x5 - 5x3 + 3x + 2This has 2 changes of sign. Therefore the number of negative roots is either 2 or 0.


THINK!

How many complex roots does the polynomial 2x5 + 8x3 + 3x - 7 have?


THINK!


1 change of sign, so 1 positive root.


THINK!


1 change of sign, so 1 positive root.

-2x5 - 8x3 - 3x - 70 changes of sign, so 0 negative roots.


THINK!


There are 5 roots altogether, only 1 of which is real. So, there are 4 complex roots.


UNIT 2, LESSON 5 DESCARTES LAW OF SIGNS. FINDING POLYNOMIALS WITH GIVEN ZEROS If we are given the...

Documents

Transcript of UNIT 2, LESSON 5 DESCARTES LAW OF SIGNS. FINDING POLYNOMIALS WITH GIVEN ZEROS If we are given the...